1
\$\begingroup\$

I'm designing the devices that should communicate via a long CAN bus (500+ meters). The power (24V) to the devices is delivered via the same bus cable (UTP5e), by the rest 3 pairs.

The limiting factor for maximum number of devices and given bus length is supply current flowing to the devices. So I'm trying to estimate the number of devices based on the current consumption of all the devices.

The main current consumer in each device is a CAN transceiver itself (NXP TJA-1042). The single device consumes up to 70mA of current when driving CAN bus to the dominant state.

  1. Lets assume I want 100 devices. If I multiply 70mA by 100 (in CAN protocol in some cases all the devices drive the bus dominant simultaneously), then I would yield the ridiculously high 7A current value! But is it legitimate to estimate the total 100 device current this way? When CAN transceiver drives the bus to a dominant state, as I understand it, all it does - its MOSFETs connect CANL to GND, and CANH to VCC and then current starts to flow via 2x120 Ohm termination resistors (60 Ohm in total in parallel). So, on the second thought, no matter how many transceivers we have, if our Vcc is 5V, we wouldn't get driving current greater then 5V/60 Ohm = 0.08A in total for all the devices (so they share this amount of current). Taking the bus resistance into account, the flowing total current would be even less. Is it a correct assumption or not?

  2. If maximum consumption of ANY CAN transceiver (not only in my example) in dominant state is determined by CAN bus topology itself and calculated as (Vcc - Vss)/Rterm + Ilogic (Ilogic - current consumed by transceiver internal logic circuits), so the minimum value in case of 5V transceivers and standard termination would be 80uA. How is it possible that existing transceivers according to datasheets have these values lower than that (usually from 20 uA min to 70uA max) even taking into account the voltage drop on MOSFETs and schottkys?

\$\endgroup\$
1
  • \$\begingroup\$ Regarding your first question, I would consider distributed power as in have multiple 24V supplies and have each board supply the transceiver from its own regulated 5V. Galvanic isolation with all transceivers supplied from a secondary 24V supply sounds problematic with so many nodes in the system. In case you'd go for that still, then maybe provide the 24V and supply GND in a separate (thick) wire and not together with the signals. \$\endgroup\$
    – Lundin
    Feb 20, 2023 at 9:46

1 Answer 1

0
\$\begingroup\$

The limiting factor may not be the supply current.

Typical electrical limit will be 112 devices on the bus for standard CAN tranceivers.

  1. The bus drive currents do not add up, because even if all devices go to domimant state, the voltage between CANH and CANL wires will be max 5V anyway, so max total of about 80mA would still flow through the resistors.

  2. I don't see what you are calculating and why, but generally the chip current consumption is defined separately from bus leakage currents. The chip will consume some amount of supply current. The bus currents are just calculated as extra load on bus.

However, since you intend to have a 500+ meter bus with 100 devices, it means you are limited with 100 kbps signaling rate. Your Cat5 cable has 100 ohm characteristic impedance, so it will load the tranceivers more, and you will get reflections if you have 120 ohm terminations on a 100 ohm bus.

The CAN tranceiver you are talking about is intended for 5 Mbps rate, so it will try to drive the bus with too fast slew rate. Just note that even for 1 Mbps rate, the bus specifications are limited to 40 meters and 30 nodes.

\$\endgroup\$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.